To show this, first note that $u\ast v$ is radially symmetric and that its radial part $\kappa_n$ solves, for every function $\varphi$,
$$
\int\varphi(r^2)\kappa_n(r)\,\mathrm dr=\int\varphi(|x|^2)\,(\mu\ast\nu)(\mathrm dx)=\iint\varphi(|as+bt|^2)\,a^{n-1}\sigma_{n-1}(\mathrm ds)b^{n-1}\sigma_{n-1}(\mathrm dt).
$$
By rotational invariance, this is
$$
\int\varphi(|as+b|^2)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds)=\int\varphi(a^2+b^2+2ab\langle s,1\rangle)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds).
$$
One knows that
$$
\sigma_{n-1}(\mathrm ds)=\sin^{n-2}\phi_1\cdot\sin^{n-3}\phi_2\cdots\sin\phi_{n-2}\cdot\mathrm d\phi_1\cdots\mathrm d\phi_{n-1},
$$
for some angles $\phi_k$ independent and uniform on $[0,\pi]$ except $\phi_{n-1}$ which is uniform on $[0,2\pi]$. Hence, $\langle s,1\rangle=\cos\theta_1$ and
$$
\int\varphi(r^2)\kappa(r)\,\mathrm dr=c_n\int_0^\pi\varphi(a^2+b^2+2ab\cos\theta)\,(ab)^{n-1}\sin^{n-2}\theta\cdot\mathrm d\theta.
$$
Finally, the change of variables $r^2=a^2+b^2+2ab\cos\theta$ in the last integral yields the result.